This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Problem 23.80 A. If a spherical raindrop of radius 0.650 mm carries a charge of 1.70pC uniformly distributed over its volume, what is the potential at its surface? (Take the potential to be zero at an infinite distance from the raindrop.)? Since the potential is zero at infinite distance from the raindrop we only need to consider the potential of the drop itself. Since it is spherically symmetric we know that the potential will be of the same form as that of a point charge located at the origin. This point charge will be equal to the total charge of the raindrop. Answer in V. V = q 4 ΠΕ o r 1.7 * 10 12 H 4 Π L I 8.85 * 10 12 M I 6.5 * 10 4 M 23.517 B.Two identical raindrops, each with radius and charge specified in part (A), collide and merge into one larger raindrop. What is the radius of this larger drop, if its charge is uniformly distributed over its volume?? Take V and R to be the volume and radius of the two original raindrops and Vnew and Rnew to be the volume of the new raindrop created from the collision of the two original ones. We can then set Vnew = 2V, substitute in the following expression for volume in terms of radius and solve for Rnew in terms of R Answer in m. V = 4 Π R 3 3 , Vnew = 4 Π Rnew 3 3 2 V = Vnew fi 2 R 3 = Rnew 3 fi 2 3 R = Rnew 2 3 I 6.5 * 10 4 M 0.000818949 C.What is the potential at its surface, if its charge is uniformly distributed over its volume?? We can use the same equation and setup as in part A but we must use Rnew for r and double q to reflect the increase in charge from the combined raindrops. Answer in V. V = 2 q 4 ΠΕ o K 2 3 r O 1.7 * 10 12 2 H 4 Π L I 8.85 * 10 12 M K 2 3 O I 6.5 * 10 4 M 37.331 Potential of a Charged Disk A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in the xy plane. Throughout this problem, you may use the variable k in place of 1 4 ΠΕ A. What is the electric potential V(z) on the z axis as a function of z , for z > 0 ? Since we can add up the individual contributions to the potential from each piece of the charged disk to find the total potential at a given location we will find V(z) by integrating over the entire disk. This is most easily accomplished in polar (cylindrical) coordinates. For and arbitrary point{0, 0, z} on the zaxis and an arbitrary point {r, Θ , 0} in the disk we can write the distance between them as z 2 + r 2 . This yields the contribution to the potential, dV from one point in the disk. dV = Σ k z 2 + r 2 where Σ is the surface charge density of the disk given Q Π a 2 . Integrating in polar coordinates and making sure to use the approriate area element dA = r dr d Θ . Note that the symmetry of the problem about the zaxis demands an answer that is independent of Θ , which is confirmed by the solution of the integral....
View
Full
Document
This note was uploaded on 09/14/2010 for the course UNKNOWN 09 taught by Professor Menke during the Spring '10 term at UC Merced.
 Spring '10
 Menke
 Charge

Click to edit the document details